Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int grid[101][101];
int grid_test[101][101];
int row[101];
int col[101];
int row_sol[101];
int col_sol[101];
int main() {
  int n, m;
  int max = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> grid[i][j];
      if (grid[i][j] > max) max = grid[i][j];
    }
  bool found = false;
  long long min_moves = 100000000000;
  for (int k = 0; k <= max; k++) {
    for (int i = 0; i < n; i++) row[i] = 0;
    for (int j = 0; j < m; j++) col[j] = 0;
    long long moves = k;
    bool gotoMainLoop = false;
    row[0] += k;
    for (int i = 0; i < n && !gotoMainLoop; i++) {
      for (int j = 0; j < m && !gotoMainLoop; j++) {
        grid_test[i][j] = 0;
        if (i == 0) {
          if (k > grid[i][j]) {
            gotoMainLoop = true;
            break;
          }
          grid_test[i][j] += k;
        }
      }
    }
    for (int j = 0; j < m && !gotoMainLoop; j++) {
      int rem;
      if (grid_test[0][j] < grid[0][j]) {
        rem = grid[0][j] - grid_test[0][j];
        col[j] += rem;
        moves += rem;
        for (int i = 0; i < n; i++) {
          if (grid_test[i][j] + rem <= grid[i][j])
            grid_test[i][j] += rem;
          else {
            gotoMainLoop = true;
            break;
          }
        }
      }
    }
    for (int i = 0; i < n && !gotoMainLoop; i++) {
      int rem;
      rem = grid[i][0] - grid_test[i][0];
      row[i] += rem;
      moves += rem;
      for (int j = 0; j < m && !gotoMainLoop; j++) {
        if (grid_test[i][j] + rem != grid[i][j]) {
          gotoMainLoop = true;
          break;
        }
      }
    }
    if (!gotoMainLoop and moves < min_moves) {
      min_moves = moves;
      for (int i = 0; i < n; i++) row_sol[i] = row[i];
      for (int j = 0; j < m; j++) col_sol[j] = col[j];
    }
  }
  if (min_moves == 100000000000)
    cout << -1 << endl;
  else {
    cout << min_moves << endl;
    for (int i = 0; i < n; i++)
      for (int t = 0; t < row_sol[i]; t++) cout << "row " << i + 1 << endl;
    for (int j = 0; j < m; j++)
      for (int t = 0; t < col_sol[j]; t++) cout << "col " << j + 1 << endl;
  }
  return 0;
}